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3 years ago

What is not changed when work is done by a machine?

Physics

2 answers:

den301095 [7]3 years ago

4 0

B) The amount of work done

maks197457 [2]3 years ago

3 0

Your answer is B
hope this helps
:):):):):):):)

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Help wanted on this question.

IceJOKER [234]

9.807 newts so c 10!!!

6 0

3 years ago

Read 2 more answers

You are explaining why astronauts feel weightless while orbiting in the space shuttle. Your friends respond that they thought gr

Yuri [45]

Answer:

The answer is "83.1%".

Explanation:

Given:

$\text{Mass of the earth}\ (M_E) = 5.97 \times 10^{24}\ \ kg\\\\\text{Radius of the earth}\ (R_E) = 6380 \ km = 6.38 \times 10^{6}\ \ m\\\\\text{acceleration of gravity}\ (g_E) = 608 \ \ km= 608,000 \ \ m \\\\G= 6.67 \times 10^{-11} \ \ \frac{N \cdot n^2}{kg^2}$

Using formula:

$\to g_E = G \frac{M_E}{(R_E +h)^2}$

$\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6.38 \times 10^{6}+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km} =6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,988,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{4.88 \times 10^{13}}\\\\$

$\to g_{608 \ km}=6.67\times10^{-24}\frac{5.97 \times 10^{24}}{4.88}\\\\\to g_{608 \ km}=6.67\times \frac{5.97}{4.88}\\\\\to g_{608 \ km}=6.67\times 1.22336066\\\\\to g_{608 \ km}= 8.15 \ \frac{m}{s^2}$

Calculating the gravity on the Earth’s surface:

$\to \frac{g_{608 \ km} }{ g_{\ earth \ surface}}$ $= \frac{8.15}{9.8} \times 100=83.1 \%$

8 0

3 years ago

This is a waste of time.

slega [8]

Just gonna grab some points by answering this. Have a good day!

6 0

4 years ago

A cup containing 200 g of hot water is taken off the stove placed on the kitchen table. Initially the water is at 75Degree C. bu

8090 [49]

Answer:

ΔH = -45.1872 kJ , where negative sign signifies heat loss.

Q = ΔH = -45.1872 kJ , where negative sign signifies heat loss.

S system = -0.141 kJ/K

S surroundings = 0.1536 kJ/K

S universe = -0.141 kJ/K + 0.1536 kJ/K = 0.0126 kJ/K

Explanation:

Given:

Cp = 4. 184 J/(mole. K)

T₁ = 75 ⁰C

T₂ = 21 ⁰C

Mass of water = 200 g = 0.2 kg

Since,

$\Delta H=m\times C\times (T_f-T_i)$

ΔH = 0.2*4.184*(21-75) kJ

ΔH = -45.1872 kJ , where negative sign signifies heat loss.

Since the process is at constant pressure

Q = ΔH = -45.1872 kJ , where negative sign signifies heat loss.

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

T₁ = 75 ⁰C = 348 .15 K

T₂ = 21 ⁰C = 294.15 K

The entropy of the water is given by:

S = m×Cp×ln(T₂ /T₁)

S = 0.2*4.184*ln(294.15/348.15)

S system = -0.141 kJ/K

The heat gain by surroundings

dQ = -Qreaction = 45.1872 kJ

The entropy change of surroundings is

S surr = dQ/T₂ = 45.1872/294 .15

S surr = 0.1536 kJ/K

The entropy of universe is the sum total of the entropy of the system and the surroundings and thus,

S universe = Ssys + Ssurr

S universe = -0.141 kJ/K + 0.1536 kJ/K = 0.0126 kJ/K

4 0

3 years ago

A certain cable of an elevator is designed to exert a force of 4.5x 104 N. If the maximum acceleration that a loaded car can wit

meriva

Answer: 3,383.5 kg

Explanation:

from the question we were given the following

tension (T) = 4.5 x 10^4 N

maximum acceleration (a) = 3.5 m/s^2

acceleration due to gravity (g) = 9.8 m/s^2 ( it's a constant value )

mass of the car and its contents (m) = ?

we can get the mass of the car and it's contents from the formula for tension which is T = ma + mg

T = m (a + g)

therefore m = T / (a+g)

m = (4.5 x 10^4 / ( 3.5 + 9.8 )

m = 3,383.5 kg

5 0

3 years ago